Verify if the product (3-2i)(3+2i) is complex?
The non-real component of the product is (-2i * 3) +(3 * 2i)
This is -6i + 6i =0
Therefore the product is real number. Real component being 9 +4 =13.
We know that multiplying a complex number by it's conjugate the result is a real number.
The special product returns a difference of two squares.
`(a - bi)(a + bi) = a^2 + abi - abi - b^2*i^2`
But `i^2` = -1
`(a - bi)(a + bi) = a^2 +b^2`
Let a = 3 and b = 2i
`(3 - 2i)(3 + 2i) = 3^2 +2^2`
(3 - 2i)(3 + 2i) = 9 + 4
(3 - 2i)(3 + 2i) = 13
Therefore, the given product is a real number: (3 - 2i)(3 + 2i) = 13.